# PDE for Heat Diffusion Equation

• ZedCar
In summary, the heat diffusion equation is a mathematical model that describes the flow of heat in a solid material due to a difference in temperature. It has two variables, time and space, which represent the rate of change of temperature and the location within the material. This equation represents the physical process of heat transfer and can be solved using numerical methods. It has many applications in fields such as engineering, physics, and chemistry, including heat transfer analysis and temperature prediction.
ZedCar

## Homework Statement

The one-dimensional heat diffusion equation is given by :

∂t(x,t)/∂t = α[∂^2T(x,t) / ∂x^2]

where α is positive.

Is the following a possible solution? Assume that the constants a and b can take any positive value.

T(x,t) = exp(at)cos(bx)

## The Attempt at a Solution

T(x,t) = exp(at)cos(bx)

∂T/∂t = a exp(at) cos(bx)
∂^2 T/∂x^2 = -b^2 exp(at) cos(bx)

As a, b and α are all positive this cannot be a solution.

A friend and I were working on this and got the answer above. Though as it was about a week ago I can't exactly remember what we've done here.

In the first step I think we've differentiated exp(at) with respect to t, treating the cos(bx) as a constant multiplier, which become a exp(at) cos(bx).

Is that all correct above?

Then we differentiated again with respect to t.

So wouldn't that become;

∂^2 T/∂x^2 = a^2 exp(at) cos(bx)

Why did we previously get ∂^2 T/∂x^2 = -b^2 exp(at) cos(bx)

Thanks!

ZedCar said:

## Homework Statement

The one-dimensional heat diffusion equation is given by :

∂t(x,t)/∂t = α[∂^2T(x,t) / ∂x^2]
That first "t" should be "T". I realize this is a typo.

where α is positive.

Is the following a possible solution? Assume that the constants a and b can take any positive value.

T(x,t) = exp(at)cos(bx)

## The Attempt at a Solution

T(x,t) = exp(at)cos(bx)

∂T/∂t = a exp(at) cos(bx)
∂^2 T/∂x^2 = -b^2 exp(at) cos(bx)

As a, b and α are all positive this cannot be a solution.

A friend and I were working on this and got the answer above. Though as it was about a week ago I can't exactly remember what we've done here.

In the first step I think we've differentiated exp(at) with respect to t, treating the cos(bx) as a constant multiplier, which become a exp(at) cos(bx).

Is that all correct above?

Then we differentiated again with respect to t.
Why?

So wouldn't that become;

∂^2 T/∂x^2 = a^2 exp(at) cos(bx)
Yes, but the second derivative with respect to t is irrelevant. Your differential equation has only the first derivative with respect to t.

Why did we previously get ∂^2 T/∂x^2 = -b^2 exp(at) cos(bx)
Because that is the correct second derivative with respect to x, not t.

Thanks!
The first derivative with respect to t is a exp(at()cos(bx). If the differential equation ∂T/∂t= a∂^2T/∂x^2 were satisfied by exp(at)cos(bx) you would have to have
aexp(at)cos(bx)= -ab^2exp(at)cos(bt) which would imply a= -ab^2 which in turn gives 1= -b^2 so b^2= -1 which is impossible for real valued b.

Cheers for that! I see what I was doing wrong - getting x and t confused...

Another possible solution we investigated for the same heat diffusion equation was;

T(x,t) = exp(-at + ibx)The PDEs being;

∂T/∂t = -a exp(-at + ibx)

∂^2 T / ∂x^2 = - b^2 exp(-at + ibx)So the possible solution is possible if a = b^2 α
I understand all the above now.

Though apparently T(x, t) is complex in this instance. How is this apparent?

ZedCar said:

## Homework Statement

The one-dimensional heat diffusion equation is given by :

∂t(x,t)/∂t = α[∂^2T(x,t) / ∂x^2]

where α is positive.

Is the following a possible solution? Assume that the constants a and b can take any positive value.

T(x,t) = exp(at)cos(bx)

## The Attempt at a Solution

T(x,t) = exp(at)cos(bx)

∂T/∂t = a exp(at) cos(bx)
∂^2 T/∂x^2 = -b^2 exp(at) cos(bx)

As a, b and α are all positive this cannot be a solution.

A friend and I were working on this and got the answer above. Though as it was about a week ago I can't exactly remember what we've done here.

In the first step I think we've differentiated exp(at) with respect to t, treating the cos(bx) as a constant multiplier, which become a exp(at) cos(bx).

Is that all correct above?

Then we differentiated again with respect to t.

So wouldn't that become;

∂^2 T/∂x^2 = a^2 exp(at) cos(bx)

Why did we previously get ∂^2 T/∂x^2 = -b^2 exp(at) cos(bx)

Thanks!

Be very glad that a, b > 0 does not give a solution, since such a "solution" would violate some basic laws of Physics: warm regions (where cos(bx) > 0) would get hotter, and cool regions (where cos(bx) < 0) would get colder, so heat would flow from cold to hot without any external influence---violating Thermodynamics. Also, warm regions would eventually get hotter than the hottest star, and cold regions would plunge well below absolute zero.

Look instead, at the case a < 0, b > 0. Does that behave properly?

RGV

## 1. What is the heat diffusion equation?

The heat diffusion equation, also known as the heat equation, is a partial differential equation (PDE) that describes the flow of heat in a solid material. It is a mathematical model that represents the transfer of thermal energy through a medium due to a difference in temperature.

## 2. What are the variables in the heat diffusion equation?

The heat diffusion equation has two variables: time (t) and space (x). Time represents the rate of change of temperature, while space represents the location within the material where the temperature is being measured.

## 3. What is the physical meaning of the heat diffusion equation?

The heat diffusion equation represents the physical process of heat transfer through a material. It describes how the temperature of a material changes over time due to the flow of thermal energy.

## 4. How is the heat diffusion equation solved?

The heat diffusion equation can be solved using various numerical methods such as the finite difference method, finite element method, or the method of lines. These methods discretize the equation and solve for the temperature at each point in space and time.

## 5. What are the applications of the heat diffusion equation?

The heat diffusion equation has numerous applications in various fields such as engineering, physics, and chemistry. It is commonly used in heat transfer analysis, temperature prediction in materials, and modeling of thermal systems such as heat exchangers and electronic devices.

• Calculus and Beyond Homework Help
Replies
0
Views
537
• Calculus and Beyond Homework Help
Replies
12
Views
556
• Calculus and Beyond Homework Help
Replies
5
Views
694
• Calculus and Beyond Homework Help
Replies
2
Views
768
• Calculus and Beyond Homework Help
Replies
8
Views
465
• Calculus and Beyond Homework Help
Replies
3
Views
1K
• Calculus and Beyond Homework Help
Replies
5
Views
430
• Calculus and Beyond Homework Help
Replies
18
Views
2K
• Calculus and Beyond Homework Help
Replies
9
Views
1K
• Calculus and Beyond Homework Help
Replies
7
Views
541